『北區高中物理科學人才培育』計畫高二物理期末考試卷

『北區高中物理科學人才培育』計畫高二物理期末考試卷 1 The electric potential between the electrodes of a vacuum tube diode is given by V(x) = cx3 , where x is the distance from the cathode and C is a constant. Assume that the distance between the cathode and anode is 13.0 mm and the potential difference between electrodes is 240V (a) Determine the value of C. (10%) (b) Obtain a formula for the electric field between the electrodes as a function of x. (10%). 2 A source with emf E. and internal resistance r is connected to an external circuit. (a) Show. that the power output of the source is maximum when the current in the circuit is one-half the short-circuit current of the source. (10%) (b) If the external circuit consists of a resistance R, show that the power output is maximum when R = r and that the maximum power is E 2/4r. (10%) 3 An electron follows a helical path in a uniform magnetic field given by B = (20i - 50j - 30k) mT. At time t = 0, the electron’s velocity is given by v = (20i - 30j + 50k) m/s. (a) What is the angle φbetween v and B ? What is the (b) radius, (c) pitch(螺距) of the helical path? (20%). 4 二線圈配置如右圖，其自感分別為 L1 及 L2、互感為 M；試證 當其為串聯時之等效電感為 (20%). Leq  L1  L2  2M . 5 二玻璃板於一端接觸，另端分離如右圖。當波長 600nm 的光垂 直入設於上板時，你會看到上板上有 9 條暗紋和 8 條亮紋。如 果你把二板在分離端的距離再增加 600nm，此時上板上會有幾 條暗紋? (20%).

1 Sol: (a) V  Cx4/3 C  V/x4/3  240 V/(130 103 m)4/3  785 104V/m4/3. (b). Ex  . V 4   Cx1/3  (105  105V/m4/3 ) x1/3 x 3. The minus sign means that Ex is in the.  x-direction,. which says that E points from the. positive anode toward the negative cathode. 2 Sol: (a). P   I  I 2r , so. dP 1 1    2 Ir  0 for maximum power output and I P max   Ishort circuit . dI 2r 2. (b) For the maximum power output of part (a), I . . 1 . r  R  2r and R  r. 2r.  .   P  I 2R    r  4r  2r  2. Then,.  rR. 2. 3 Sol: (a) v∙B=vBcosa= cos(2/19) = 84. (b) We find v = v sin = 61.3 m/s, so r = mv /eB = 5.7 nm. (c) v∥ = v cos = 6.5 m/s, so p = 2mv∥/eB = 40.7 nm. 4 Sol: L1. di1 di di di di  L2 2  M 21 1  M12 2  Leq . dt dt dt dt dt. But i  i1  i2 . di di1 di2   and M12  M 21  M , dt dt dt. so ( L1  L2  2M ). di di  Leq and Leq  L1  L2  2M . dt dt. 5 Sol: By the condition m = 2y where y is the thickness of the air-film between the plates directly underneath the middle of a dark band), the edge of the plates (the edge where they are not touching) are y = 8/2 = 2400 nm apart (where we have assumed that the middle of the ninth dark band is at the edge). Increasing that to y' = 3000 nm would correspond to m' = 2y'/ = 10 (counted as the eleventh dark band, since the first one corresponds to m = 0). There are thus 11 dark fringes along the top plate..